Answer:
1. 9.90 x 10^14 Hz
Explanation:
The work function is the minimum energy required to eject electron's from the surface of the metal. The frequency of photons that are capable of doing this is called the threshold frequency. Given the work function, the threshold frequency can be calculated in the following way
Work function = Planck's constant * frequency
6.56 x 10^-19 J = 6.62607004 × 10-34 m²-kg/s * f
f = 9.90 x 10^14 Hz
A buffer of pH 9.24 is to be prepared from a weak acid and its salt. The best acid from which to prepare the buffer is
(A) phthalic acid, K1 = 1.3 x 10-3 (1st ionization)
(B) hydrohen phthalate, K2 = 3.9 x 10-5
(C) arsenious acid, Ka = 6 x 10-10
(D) formic acid, Ka = 1.8 x 10-5
(E) phenol, Ka = 1.3 x 10-10
Answer:
(C) arsenious acid, Ka = 6 x 10⁻¹⁰
Explanation:
A buffer is prepared by a weak acid and the conjugate base coming from its salt. Its function is to resist abrupt changes in pH when an acid or a base are added. The best working range of a buffer is in the range of pKa ± 1. Let's consider the 5 options and their pKa (pKa = -log Ka).
(A) phthalic acid, K1 = 1.3 x 10⁻³ (1st ionization) pKa = 2.9
(B) hydrogen phthalate, K2 = 3.9 x 10⁻⁵ pKa = 4.4
(C) arsenious acid, Ka = 6 x 10⁻¹⁰ pKa = 9
(D) formic acid, Ka = 1.8 x 10⁻⁵ pKa = 4.7
(E) phenol, Ka = 1.3 x 10⁻¹⁰ pKa = 9.8
The acid whose pKa is closer to the desired pH is arsenious acid. Its working range of pH is 8 - 10. In the second place, phenol could work as a buffer system since the working pH range is 8.8 - 10.8.
Which of the following metal ions is most likely to exist as a stable eight-coordinate complex with fluoride? Cu2+ Cu+ Fe3+ Ag3+ U4+
Answer:
U⁴⁺
Explanation:
Fluorine stabilizes metals in higher oxidation states with high M:F ratios.
Hard acids prefer to bind to hard bases.
U⁴⁺ is the hardest acid in the list and F⁻ is a hard base, so an eight-coordinate complex of U⁴⁺ is the most stable.
The metal ion that can most adequately exist in the form of 8-coordinate complex would be:
d). U⁴⁺
'Coordination Complex' is characterized as either an ion or a compound that consists of the main metallic atom that is linked through interrelated bonds employing a fixed number of nearby encircling atoms.In the given question, the metal ion U⁴⁺ would most adequately retain in the form of 8 coordinate complexes because it allows metals to sustain in a greater rate of oxidation having a greater ratio of M:F. Since hard bases are defined by stronger acids, U⁴⁺ would be most appropriate as it is a hard acid while F⁻ is a strong base which makes it reliable.
Thus, option d is the correct answer.
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Which element is reduced in the following reaction? Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O Which element is reduced in the following reaction? Fe2S3 + 12HNO3 2Fe(NO3)3 + 3S + 6NO2 + 6H2O H N O NO2 S
Answer: The nitrogen atom is getting reduced.
Explanation:
Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
The given chemical equation follows:
[tex]Fe_2S_3+12HNO_3\rightarrow 2Fe(NO_3)_3+3S+6NO_2+6H_2O[/tex]
On reactant side:
Oxidation state of iron atom = +3
Oxidation state of nitrogen atom = +5
Oxidation state of sulfur atom = -2
On product side:
Oxidation state of iron atom = +3
Oxidation state of sulfur atom = 0
Oxidation state of nitrogen atom in [tex]NO_2[/tex] = +4
Oxidation state of nitrogen atom in [tex]Fe(NO_3)_3[/tex] = +5
As, the oxidation state of nitrogen atom in [tex]NO_2[/tex] is decreasing from +5 to +4. So, it is getting reduced.
And, oxidation state of sulfur atom is increasing from -2 to 0. So, it is getting oxidized.
Hence, the nitrogen atom is getting reduced.
The element that gets reduced in the chemical reaction is nitrogen.
What is reduction?Reduction in chemistry means when an atom accepts electrons to become more negatively charged.
According to this question, the following chemical reaction is given:
Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O
On reactant side:
Oxidation state of iron atom = +3Oxidation state of nitrogen atom = +5Oxidation state of sulfur atom = -2On product side:
Oxidation state of iron atom = +3Oxidation state of sulfur atom = 0Oxidation state of nitrogen atom in = +4As indicated above, the oxidation number of nitrogen reduces from +5 to +4, hence, it is the element that gets reduced.
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Which of the following statements is false? The mean free path of a molecule depends on the size of the molecule. ODalton's law of partial pressures indicates that in gaseous mixtures at low pressure each kind of molecule behaves independently of the others O At a given temperature, for a given gas, every molecule has the same speed Collisions of molecules with the container walls give rise to the gas pressure O At high pressures different gases give different values for the ratio PV/nRT.
Answer:
False statement is
At a given temperature, for a given gas, every molecule has the same speed
Explanation:
While checking each and every statement given
The mean free of a molecule actually depends on the size of the molecule because the mean free path is defined as the average distance between two successive collisions of the gas molecules
If the size of the molecule is more, the average distance between two successive collisions decrease and as a result the mean free path of the molecule decreases
Dalton's law of partial pressures is applicable for only ideal gases which means we are assuming that the size of the molecule of a gas is negligible and there are no intermolecular forces of attraction
These two assumptions gets applied at high temperature and low pressure
So Dalton's law of partial pressures tells us that total pressure of the gas is equal to the sum of the partial pressures of the individual gas components
∴ It explains the independent nature of the gas molecule
At a given temperature, for a given gas, all molecules of the gas do not have same speed but overall the average speed of the gas remains same because speed of each molecule of a gas depends on the collision with other molecules of the gas and as the collisions can't be the same therefore molecules of a gas have different speeds
Actually pressure is generated by the collisions of molecules with the container walls because when the gas molecules collide with the container they generate a force which in turn produce the pressure
At high pressure gas do not tend to behave ideally as there will be intermolecular forces and we will write the ratio of PV/nRT as Z which is the compressibility factor of a gas and it will be different for different gases as different gases has different intermolecular forces of attraction
Imagine that you have a 7.00 L gas tank and a 3.00 L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 145 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
Final answer:
To run out of oxygen and acetylene at the same time, the acetylene tank should be filled to a pressure of 62.14 atm.
Explanation:
To ensure that you run out of oxygen and acetylene at the same time, the number of moles of oxygen and acetylene consumed must be equal. The volume ratio of oxygen and acetylene gases is 7.00 L : 3.00 L, which simplifies to 7 : 3. Therefore, the pressure ratio of oxygen to acetylene should also be 7 : 3. Given that the oxygen tank is filled to a pressure of 145 atm, the acetylene tank should be filled to a pressure of (145 atm) x (3/7) = 62.14 atm.
Kemmi Major does some experimental work on the combustion of sucrose: C12H22O11(s) 12 O2(g) → 12 CO2(g) 11 H2O(g) She burns a 0.05392 g pellet of sucrose in a bomb calorimeter with excess oxygen. She determines the qrxn to be –916.6 J for the reaction. Calculate the ∆H value for the combustion reaction. (Round the answer to 3 significant digits, units of kJ, pay attention to positive or negative.
Answer: [tex]5.81\times 10^6J/mol[/tex]
Explanation:
Heat of combustion is the amount of heat released when 1 mole of the compound is completely burnt in the presence of oxygen.
[tex]C_{12}H_{22}O_{11}(s)+12O_2\rightarrow 12CO_2(g)+11H_2O(g)[/tex]
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.05392g}{342g/mol}=1.577\times 10^{-4}moles[/tex]
Thus [tex]1.577\times 10^{-4}moles[/tex] of sucrose releases = 916.6 J of heat
1 mole of sucrose releases =[tex]\frac{916.6}{1.577\times 10^{-4}}\times 1=5.81\times 10^6J[/tex] of heat
Thus ∆H value for the combustion reaction is [tex]5.81\times 10^6J/mol[/tex]
The ∆H for the combustion reaction of 0.05392 g of sucrose, given that the qrxn is -916.6 J, is calculated to be approximately -58364.2 kJ. This value is negative indicating an exothermic reaction where heat is released.
Explanation:In this problem, we are tasked with finding the enthalpy change, or ∆H, for the combustion reaction of sucrose. The heat of the reaction, qrxn, is given as -916.6 J for a 0.05392 g pellet of sucrose.
Firstly, we have to convert the grams of sucrose to moles. The molecular weight of sucrose, C12H22O11, is about 342.3 g/mol so the moles of sucrose burned would be 0.05392 g ÷ 342.3 g/mol = 0.000157 mol.
Next, the change in enthalpy per mole of sucrose can be calculated using ∆H = qrxn / moles of sucrose which gives us -916.6 J / 0.000157 mol = -58364152.2293 J/mol. Since the question requires the answer in kJ, we must divide this by 1000 to convert J to kJ. Hence, ∆H = -58364.1523 kJ/mol.
It is important to note that ∆H is negative because the reaction is exothermic, meaning heat is released during the combustion of sucrose. Therefore, the ∆H for the combustion reaction of sucrose is approximately -58364.2 kJ.
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The amount of I3-(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O3^2-(aq) (thiosulfate ion).
The determination is based on the net ionic equation
2S2O3^2-(aq)+I3^-(aq) ---> S4O6^2-(aq)+31^-(aq)
Given that it requires 35.8 mL of 0.350 M Na2S2O3(aq) to titrate a 30.0-mL sample of I3^-(aq) calculate the molarity of I3^-(aq) in the solution
Answer:
molarity of I3^-(aq) in the solution = 0.21 M
Explanation:
We are given the balanced chemical reaction and the volume and molarity of the Na₂S₂O₃ so we can calculate the moles of the thiosulfate that were required, and then can calculate the molarity the I₃⁻ by the definition of molarity.
First lets convert the volume of Na₂S₂O₃ to liters:
35.8 mL x 1 L/1000 mL = 0.0358 L
# moles Na₂S₂O₃ = 0.350 mol/L x 0.0358 L = 0.0125 mol
From the stoichiometry of the reaction we know 2 mol Na₂S₂O₃ s react with 1 mol I₃⁻ , therefore mol of I₃⁻ will be given by
1 mol I₃⁻ / 2 mol Na₂S₂O₃ x 0.0125 mol Na₂S₂O₃ = 0.0063 mol I₃⁻
and its molarity is:
0.0063 mol I₃⁻ / 0.030 L = 0.21 M
Investigating iron thiocyanate requires several chemicals which must be used with caution. Using knowledge of these chemicals, select whether the following statements are true or false. When handling HNO3, it is important to wash your hands frequently since HNO3 is corrosive and can cause burns. To dilute an acid, always add water to add while slowly stirring. Fe(NO3)3 should be handled with caution because it is an oxidizer and can irritate the skin. KSCN is completely nontoxic If KSCN is strongly heated it will cause the evolution of cyanide gas. KSCN, if combined with a strong base, will also cause the evolution of cyanide gas.
Answer:
1. False
2. False
3. True
4. False
5. False
6. False
Explanation:
1. Because HNO₃ is corrosive when handling it is important to use gloves and also protection glasses. If it touches the skin, then it's important to wash the area affected.
2. The dilute of acid is exothermic, so it releases a huge amount of heat. For precaution, the acid must be added at the water slowly stirring, so it can be controlled.
3. Fe(NO₃)₃ is an oxidizer and in contact with the skin it can cause irritation, so it must be handled carefully and with protection.
4. KSC can irritate the eyes, so it's toxic.
5. KSCN is not inflammable and it's not combustible, so when heated it will only change the state for gas, but it will not cause the evolution of cyanide gas.
6. If KSCN reacts with a strong base, it will dissociate, and the ions SCN⁻, which can't react with the OH⁻ ions of the base. So, it will only cause the evolution of cyanide gas if it reacts with a strong acid.
The answers have been given according to which is True or False. It explores the science of chemicals, especially Nitric Acid.
What is a Chemical Reaction?
Two substances will achieve a reaction has occurs when the molecular or ionic structure of a substance is rearranged to create a new form.
The statements thus are:
1) True.
2) False.
3) True.
4) False
5) False
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Calculate the standard molar enthalpy of formation, in kJ/mol, of NO(g) from the following data:
N2 + 2NO2 = 2NO2 ΔH 66.4kJ @298k
2NO + 02 =2NO2 ΔH -114.1kJ @298k
Answer:
The standard molar enthalpy of formation of 1 mole of NO gas is 90.25 kJ/mol.
Explanation:
We have :
[tex]N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ[/tex]
[tex]2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o_{2} = -114.1 kJ[/tex]
To calculate the standard molar enthalpy of formation
[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]...[3]
[1] - [2] = [3] (Hess's law)
[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]
[tex]\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2} [/tex]
[tex]\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ[/tex]
According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide.
So, the standard molar enthalpy of formation of 1 mole of NO gas :
[tex]\Delta H^o_{f,NO}=\frac{\Delta H^o_{3}}{2 mol}[/tex]
[tex]\Delta H^o_{f,NO}=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol[/tex]
To calculate the standard molar enthalpy of formation of NO(g), the given reactions are used by reversing the second reaction and adding its enthalpy change to the first reaction's ΔH. Dividing the total by 2 gives 90.25 kJ/mol for the formation of 1 mole of NO(g) from its elements.
Explanation:To calculate the standard molar enthalpy of formation of NO(g), we can use the given enthalpy changes for the reactions provided. The two reactions are:
N2(g) + 2O2(g) → 2NO2(g), ΔH = +66.4 kJ2NO(g) + O2(g) → 2NO2(g), ΔH = -114.1 kJWe want to find the enthalpy change for the formation of NO(g) from its elements, which is N2(g) and O2(g). This is the standard molar enthalpy of formation for NO(g). We can set up an enthalpy diagram to help visualize the process.
Steps to Calculate the Standard Molar Enthalpy of Formation:Reverse the second reaction to find the enthalpy change for breaking down 2NO2 into 2NO and O2, which will be +114.1 kJ (reversing a reaction changes the sign of ΔH).Add this value to the first reaction's ΔH, to get the total enthalpy change for the formation of 1 mole of NO(g) from its elements.Working through the math:
ΔHf[NO] = (66.4 kJ + 114.1 kJ) / 2
ΔHf[NO] = 180.5 kJ / 2
ΔHf[NO] = 90.25 kJ/mol
Therefore, the standard molar enthalpy of formation of NO(g) is 90.25 kJ/mol.
A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in a 25.0 L container at 713 K. At equilibrium, 72.6 g H2S (g) is present? What is the value of K at 713 Kelvin for the following reaction?
Answer: The value of [tex]K_c[/tex] is coming out to be 0.412
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For [tex]Sb_2S_3[/tex]Given mass of [tex]Sb_2S_3[/tex] = 1.00 kg = 1000 g (Conversion factor: 1 kg = 1000 g)
Molar mass of [tex]Sb_2S_3[/tex] = 339.7 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }Sb_2S_3=\frac{1000g}{339.7g/mol}=2.944mol[/tex]
For hydrogen gas:Given mass of hydrogen gas = 10.0 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol[/tex]
For hydrogen sulfide:Given mass of hydrogen sulfide = 72.6 g
Molar mass of hydrogen sulfide = 34 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol[/tex]
The chemical equation for the reaction of antimony sulfide and hydrogen gas follows:
[tex]Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)[/tex]
Initial: 2.944 5
At eqllm: 2.944-x 5-3x 2x 3x
We are given:
Equilibrium moles of hydrogen sulfide = 2.135 moles
Calculating for 'x', we get:
[tex]\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712[/tex]
Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles
Volume of the container = 25.0 L
Molarity of a solution is calculated by using the formula:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex]
The expression of [tex]K_c[/tex] for above equation, we get:
[tex]K_c=\frac{[H_2S]^3}{[H_2]^3}[/tex]
The concentration of solids and liquids are not taken in the expression of equilibrium constant.
[tex]K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412[/tex]
Hence, the value of [tex]K_c[/tex] is coming out to be 0.412
The estimated value of K for the reaction Sb2S3 (s) + 3 H2 (g) <--> 2 Sb (s) + 3 H2S (g) at 713K is approximately 1. This estimate is based on initial and equilibrium moles and concentrations of H2 and H2S.
Explanation:The reaction is Sb2S3 (s) + 3 H2 (g) <--> 2 Sb (s) + 3 H2S (g). This type of equation is a chemical equilibrium reaction. Because Sb2S3 and Sb are solids, they are not included in the equilibrium constant expression. Only H2 (g) and H2S (g) are included. To solve the problem, the initial moles of H2 and H2S need to be calculated, and the moles of H2S at equilibrium can be found. After finding the moles, they can be converted into concentrations by dividing by the volume of the container (25.0 L).
Initially, we have 1.00 kg of Sb2S3, which is excess and doesn't appear in the equilibrium equation. So, it can be ignored. Similarly, for hydrogen, 10 g converts to approximately 5 moles. Since no H2S is present initially, let's denote its moles as 0. When the equilibrium is established we know that only 72.6g of H2S is present and that is approximately 3 moles. By stoichiometry, 1 mole of H2 will form 1 mole of H2S. Therefore, 3 moles of H2 have been used to form 3 moles of H2S. Subtracting the used quantity from the total, there is approximately 2 moles of H2 left. Plugging all these into the equilibrium equation, we get K = [H2S]^3 / [H2]^3 which is approximately equal to 1.
This process illustrates the concept of chemical equilibrium and how to calculate the equilibrium constant, K, for a reaction. It emphasizes the importance of stoichiometry, the mole concept, and concentration when dealing with equilibrium reactions.
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The complete question is here:
A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in 25.0L container at 713K. At equilibrium, 72.6 g H2S (g) is present? What is the value of K at 713K for the following reaction? The value of Kp?
Sb2S3 (s) + 3 H2 (g) <--> 2 Sb (s) + 3 H2S (g)
The standard cell potential (E°cell) for the voltaic cell based on the reaction below is ________ V. Sn2+ (aq) + 2Fe3+ (aq) → 2Fe2+ (aq) + Sn4+ (aq) The standard cell potential (E°cell) for the voltaic cell based on the reaction below is ________ V. Sn2+ (aq) + 2Fe3+ (aq) 2Fe2+ (aq) + Sn4+ (aq) +1.39 +1.21 +0.46 -0.46 +0.617
The standard cell potential (E°cell) for the given electrochemical reaction is approximately +0.9 V. Here option C is correct.
Identifying the Redox Reactions and Potentials
The given balanced equation reveals two processes: an oxidation and a reduction.
Oxidation: [tex]Sn$^{2+}$[/tex] loses two electrons to become [tex]Sn$^{4+}$[/tex]. This can be represented as:
[tex]$$Sn^{2+}(aq) \rightarrow Sn^{4+}(aq) + 2e^-$$[/tex]
Reduction: [tex]Fe$^{3+}$[/tex] gains one electron to become [tex]Fe$^{2+}$[/tex]. This can be represented as:
[tex]$$Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq)$$[/tex]
Next, we need to find the standard reduction potentials (E°) for these half-reactions. Remember, we need the reverse of the oxidation reaction, so look for the potential of:
[tex]$$Sn^{4+}(aq) + 2e^- \rightarrow Sn^{2+}(aq)$$[/tex]
From the provided table, find the value for this and the given reduction reaction:
Oxidation: E° = -0.154 V
Reduction: E° = +0.771 V
Calculating the Cell Potential (E°cell)
The reduction occurs at the cathode (positive) and the oxidation at the anode (negative). Therefore, the cell potential is calculated by:
[tex]$$E^{\circ}_{cell} = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$$[/tex]
Substituting the values:
[tex]$$E^{\circ}_{cell} = (+0.771 \text{ V}) - (-0.154 \text{ V})$$[/tex]
[tex]$$E^{\circ}_{cell} \approx +0.925 \text{ V}$$[/tex]
However, the answer choices only have one decimal place, so we need to round: [tex]$$E^{\circ}_{cell} \approx \boxed{+0.9 \text{ V}}$$[/tex]. Here option C is correct.
Complete question:
The standard cell potential (E°cell) for the voltaic cell based on the reaction below is ________.
Sn2+(aq) + 2Fe3+(aq) --> 2Fe2+(aq) + Sn4+(aq)
Half-reaction E°(V)
Cr3+(aq) + 3e - --> Cr(s) - 0.74
Fe2+ (aq) + 2e - --> Fe(s) - 0.440
Fe3+(aq) + e - --> Fe2+(s) + 0.771
Sn4+(aq) + 2e - --> Sn2(aq) + 0.154
a) +0.46 V
b) +0.617 V
c) +0.9 V
d) -0.46 V
e) +1.21 V
Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presence of superheated steam. (a) Given these data, find ΔH o rxn , ΔG o rxn , and ΔS o rxn at 298 K: ΔH o f (kJ/mol) ΔG o f (kJ/mol) S o (J/mol·K) Ethylbenzene, C6H5−CH2CH3 −12.5 119.7 255 Styrene, C6H5−CH=CH2 103.8 202.5 238 ΔH o rxn ΔG o rxn ΔS o rxn kJ kJ J/K (b) At what temperature is the reaction spontaneous? °C (c) What are ΔG o rxn and K at 600.°C? ΔG o rxn K kJ/mol × 10 Enter your answer in scientific notation.
The change in enthalpy, entropy, and free energy were calculated for the dehydrogenation reaction of ethylbenzene into styrene. The reaction was found to be endothermic and results in a decrease in overall disorder. Under the given conditions, the reaction will never be spontaneous.
Explanation:The processes involved in the production of styrene from ethylbenzene are fairly complex and require knowledge of thermodynamics. We'll begin with ΔH°rxn, which is found by subtracting the enthalpy (ΔH) of the reactants from that of the products: ΔH°rxn = [ΔH°f(styrene)] - [ΔH°f(ethylbenzene)] = 103.8 kJ/mol - (-12.5 kJ/mol) = 116.3 kJ/mol. This means the reaction is endothermic, as heat is absorbed.
The change in entropy ΔS°rxn, obtained likewise, is [S°(styrene) - S°(ethylbenzene)] = (238 J/mol·K - 255 J/mol·K) = -17 J/mol·K. This indicates a decrease in disorder in the system.
With these, we can calculate the change in free energy ΔG°rxn at a given temperature (T) using the equation ΔG°rxn = ΔH°rxn - TΔS°rxn. Substituting the known values at 298 K, ΔG°rxn = 116.3 kJ/mol - (298 K)(-17 J/mol·K) = 121.2 kJ/mol, indicating a non-spontaneous reaction.
For the reaction to be spontaneous, ΔG°rxn must be less than zero. Solving for T in the above equation with ΔG°rxn = 0, yields T = ΔH°rxn / ΔS°rxn = 116.3 kJ/mol / -17 J/mol·K ≈ -6840 K. This value is negative, implying the reaction is never spontaneous under the given conditions.
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If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode? If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode?
A(aq)+1e−→A−(aq) at the cathode and B(aq)→B+(aq)+1e− at the anode.
A(aq)+1e−→A−(aq) at the anode and B(aq)→B+(aq)+1e− at the cathode
Answer:
A(aq) + 1 e⁻ → A⁻(aq) at the cathode and B(aq) → B⁺(aq) + 1e⁻ at the anode.
Explanation:
In the anode occurs the oxidation, in which the reducing agent loses electrons, and its oxidation number increases.
B(aq) → B⁺(aq) + 1e⁻
In the cathode occurs the reduction, in which the oxidizing agent gains electrons, and its oxidation number decreases.
A(aq) + 1 e⁻ → A⁻(aq)
In a voltaic cell, the anode is where oxidation occurs, and the cathode is where reduction occurs. Electrons flow from the anode (negative electrode) to the cathode (positive electrode). The anode reaction is B(aq) → B+(aq) + 1e−, and the cathode reaction is A(aq) + 1e− → A−(aq).
In a voltaic (galvanic) cell, two half-reactions occur at separate electrodes.
These reactions involve oxidation at the anode and reduction at the cathode.Let's break down the given reactions:
At the anode: The half-reaction is B(aq) → B+(aq) + 1e−. Here, oxidation occurs and electrons are released.At the cathode: The half-reaction is A(aq) + 1e− → A−(aq). Here, reduction occurs and electrons are gained.Electrode Identification
Anode: This is where oxidation takes place (B(aq) → B+(aq) + 1e−). It is the negative electrode.Cathode: This is where reduction takes place (A(aq) + 1e− → A−(aq)). It is the positive electrode.In summary, electrons flow from the anode to the cathode through an external circuit, and a salt bridge maintains charge balance.
Which of the following statements is INCORRECT? A. It is not possible to know the exact location of an electron and its exact energy simultaneously. B. The energies of an hydrogen atom's electrons are quantized. C. The wave mechanical model correctly predicts all the energy states and the orbitals available to an electron in an hydrogen atom. D. The behavior of an atom's electrons can be described by circular orbits around a nucleus.E. Electrons have both wave and particle properties.
Answer:
D. The behavior of an atom's electrons can be described by circular orbits around a nucleus.
Explanation:
Which of the following statements is INCORRECT?
A. It is not possible to know the exact location of an electron and its exact energy simultaneously. CORRECT. This is known as the Heisenberg's uncertainty principle.
B. The energies of an hydrogen atom's electrons are quantized. CORRECT. According to the modern atomic model, the energy levels are quantized, that is, they have a discrete amount of energy.
C. The wave mechanical model correctly predicts all the energy states and the orbitals available to an electron in an hydrogen atom. CORRECT. The characteristics of the energy levels are explained by Schrödinger's wave equation.
D. The behavior of an atom's electrons can be described by circular orbits around a nucleus. INCORRECT. This was postulated by Bohr's atomic model but it is now considered to be incorrect.
E. Electrons have both wave and particle properties. CORRECT. This is what De Broglie called wave-particle duality.
Statement D is incorrect because Niels Bohr's model of electrons moving in circular orbits around a nucleus has been replaced by the quantum mechanical model, which describes electrons as occupying 'orbitals' rather than specific paths.
Explanation:The statement D, 'The behavior of an atom's electrons can be described by circular orbits around a nucleus', is incorrect. This is a description of a model proposed by Niels Bohr to explain the behavior of electrons in an atom, known as the Bohr model. However, this model has been superseded by the wave mechanical model (also known as the quantum mechanical model) which states that we can't know the exact location of an electron, but we can predict where it's likely to be—an area known as an 'orbital'. Electrons do not travel in defined circular orbits as originally proposed by Bohr.
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A certain half-reaction has a standard reduction potential =E0red+0.13V . An engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least 1.10V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell.
Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have? If so, check the "yes" box and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, check the "no" box. yes, there is a minimum. =E0red V no minimum
Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have? If so, check the "yes" box and calculate the maximum. Round your answer to 2 decimal places. If there is no upper limit, check the "no" box. yes, there is a maximum. =E0red V no maximum
By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the cathode of this cell. Note: write the half reaction as it would actually occur at the cathode. e
The minimum standard reduction potential at the cathode in this galvanic cell should be at least 1.23V to provide the needed power. There is no defined maximum limit for this value. A suitable half-reaction for the cathode could involve the reduction of silver ions to silver metal.
Explanation:For a galvanic cell to work, the potential at the cathode must be greater than that at the anode. In this specific case, the reduction potential at the anode, E0anode, is given as +0.13V. The difference in potentials between the cathode and anode corresponds to the provided voltage of the cell, which is 1.10V in this particular scenario.
Using this information, the reduction potential at the cathode can be calculated using E0cell = E0cathode - E0anode formula. Substituting the known values, 1.10V = E0cathode - (+0.13V), we find that E0cathode should be at least 1.23V to provide the necessary power.
There is technically no maximum limit for E0cathode as increasing this value only increases the power provided by the cell. As for a suitable half-reaction, a reduction of silver ions (Ag+) to silver metal (Ag) having a reduction potential of +0.80V could be used:
Ag+ + e- --> Ag
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There is a minimum standard reduction potential for the cathode: 0.97 V. The maximum standard reduction potential is generally assumed to be less than +2.87 V. An example of a balanced cathode half-reaction is 2H⁺ (aq) + 2e⁻ → H₂ (g).
Yes, there is a minimum standard reduction potential for the half-reaction at the cathode. The cell potential, E°cell, is calculated as:
E°cell = E°cathode - E°anodeGiven that E°anode = -0.13 V (since it’s the reduction potential, but we use it as an oxidation potential here), and the cell must provide at least 1.10 V, we have:
1.10 V = E°cathode - (-0.13 V)1.10 V = E°cathode + 0.13 VSolve for E°cathode:
E°cathode ≥ 1.10 V - 0.13 VE°cathode ≥ 0.97 VYes, there is a maximum standard reduction potential. Practically, the maximum standard reduction potential of the cathode is determined by the potential of the reference electrode (often chosen to be +2.87 V for fluorine). Therefore, any standard reduction potential must be less than this value.
One possible balanced equation that can serve as the cathode reaction is the reduction of hydrogen ions:2H⁺ (aq) + 2e⁻ → H₂ (g); E° = 0.00 VUse the drop-down arrows to make the following statements, regarding bomb calorimetry, correct.
A bomb calorimeter is a piece of equipment designed to measure [ Select ] ["delta E", "delta H", "delta S", "C"] for reactions.
Reactions carried out in a bomb calorimeter occur at constant [ Select ] ["pressure", "volume", "mass", "temperature"] .
To calculate the heat absorbed or released by a reaction carried out in a bomb calorimeter, we use the equation: [ Select ] ["q = C * delta T", "q = m * C * delta T"]
Answer:
Answers are in the explanation
Explanation:
A bomb calorimeter is a constant volume calorimeter in which medition of change in temperature allows the medition of change in internal energy (ΔU) for chemical reactions or physical changes.
This ΔU can be converted in ΔH thus:
ΔH = ΔU + PV
That means that:
A bomb calorimeter is a piece of equipment designed to measure delta H
As there was said:
Reactions carried out in a bomb calorimeter occur at constant volume
To calculate the heat absorbed or released by a reaction carried out in a bomb calorimeter, we use the equation: q = C * delta T
Where C is the heat capacity of the calorimeter, delta T is the change in temperature and q is the heat produced in the chemical or physical process
I hope it helps!
A bomb calorimeter measures the change in internal energy (delta E) of reactions, which occur at constant volume. The heat absorbed or released is calculated using the formula q = C x delta T.
Explanation:A bomb calorimeter is a piece of equipment designed to measure delta E, or the change in internal energy, for reactions. Reactions carried out in a bomb calorimeter occur at constant volume. The heat absorbed or released by a reaction in this calorimeter is calculated using the equation q = C x delta T, where 'q' is the heat, 'C' is the heat capacity of the calorimeter, and 'delta T' is the change in temperature.
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On the basis of the Ksp values below, what is the order of the solubility from least soluble to most soluble for these compounds?
AgBr: Ksp = 5.4 x 10-13
Ag2CO3: Ksp = 8.0 x 10-12
AgCl: Ksp = 1.8 x 10-10
Ag2CO3 < AgBr < AgCl
AgBr < Ag2CO3 < AgCl
AgBr < AgCl < Ag2CO3
AgCl < Ag2CO3 < AgBr
Answer:
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Explanation:
The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:
Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).
It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:
Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.
Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that
Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋
Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12
s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶
And for AgCl
AgCl ⇄ Ag⁺ + Cl⁻
Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵
Therefore, AgCl is more soluble than Ag₂CO₃
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Answer:
5.4 x 10^-13
Explanation:
Edmentum says it is correct
demonstration of strong electrolytes, weak electrolytes, and nonelectrolytes, Professor Popsnorkle used a lightbulb apparatus that showed how much a solution conducted electricity by the brightness of the lightbulb. When pure water was tested, the bulb did not light. Then Professor Popsnorkle tested the following aqueous solutions. Which one caused the bulb to burn the brightest?
Answer:
Strong electrolyte
Explanation:
Bulb doesn't light when water is used because water doesn't have any appreciable amount of ions for the conduction of electricity.
Strong electrolytes ionize almost completely in solution, thus providing a huge amount of ions to conduct electricity. More the number of ions, more charge is carried and hence greater current and thus leading to bulb glowing brighter.
Weak electrolytes ionize partially in solution, thus providing some amount of ions for carrying charge. This results in a small amount of current going to the bulb and hence the bulb lights up with very low intensity.
Non electrolytes doesn't ionize either and hence is similar to water. These have almost zero conductivity and hence the bulb doesn't light at all.
What is electron affinity?energy change associated with the gaining of several of electrons required to complite the outermost orbital of the atom in the gaseous stateenergy change associated with the gaining of a proton by the atom in the gaseous stateenergy change associated with the gaining of an electron by the atom in the gaseous stateenergy change associated with the gaining of several electrons to complete the outermost sublevel of the atom in the gaseous state
Answer:
Energy change associated with the gaining of an electron by the atom in the gaseous state.
Explanation:
What is electron affinity?
energy change associated with the gaining of several electrons required to complete the outermost orbital of the atom in the gaseous state. FALSE. Electron affinity is associated with the gain of only 1 electron.energy change associated with the gaining of a proton by the atom in the gaseous state. FALSE. Protons cannot be gained nor lost.energy change associated with the gaining of an electron by the atom in the gaseous state. TRUE. This is the definition of electron affinity.energy change associated with the gaining of several electrons to complete the outermost sublevel of the atom in the gaseous state. FALSE. Electron affinity is associated with the gain of only 1 electron.Acetonitrile (CH3CN) is an important industrial cheical. Among other things, it Is used to make plastic moldings, which have multiple uses, from car parts to Lego bricks. Which statement below about acetonitrile is FALSE? a. Acetonitrile has 16 valence electrons in its Lewis structure. b. Acetonitrile has one triple bond. C. Acetonitrile has one pair of nonbonding electrons. d. All atoms satisfy the octet rule in acetonitrile. e. One carbon atom and the nitrogen atom have nonzero formal charges.
The molecule acetonitrile or CH3CN, indeed has 16 valence electrons and a triple bond follows its Lewis structure. However, the statement that acetonitrile has a pair of nonbonding electrons is incorrect because all 16 valence electrons are used in the molecule's bonds.
Explanation:The molecule in question is acetonitrile, or CH3CN. To analyze these statements, it's crucial to consider the Lewis structure of this molecule: the carbon atom bonded to three hydrogen atoms, that carbon bonded to another carbon atom, and that second carbon bonded to a nitrogen through a triple bond. Hydrogen has 1 valence electron, carbon has 4, and nitrogen has 5.
a. The combined total of valence electrons for acetonitrile is indeed 16 - (3x1 for hydrogen, 2x4 for carbon, and 1x5 for nitrogen).
b. There is a triple bond between the carbon and nitrogen atoms.
c. The entirety of the 16 valence electrons are used in the molecule's bonds, leaving no nonbonding electrons. This statement is false.
d. All atoms in the structure satisfy the octet rule, each having a complete outermost shell.
e. Neither carbon atom nor the nitrogen atom has a nonzero formal charge; all atoms in acetonitrile have a formal charge of zero.
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The false statement about acetonitrile (CH3CN) is that it has one pair of nonbonding electrons. All the outer shell electrons in acetonitrile are involved in bonding, hence, there are no non-bonding electrons in the molecule.
Explanation:Based on your question about acetonitrile (CH3CN), which is an important industrial chemical, the statement that is FALSE is: Acetonitrile has one pair of nonbonding electrons. When drawing the Lewis structure for acetonitrile, we find that all the outer shell electrons are involved in bonding, hence, there are no non-bonding electrons in acetonitrile. The molecule's other properties mentioned in the options are correct: it has 16 valence electrons, a triple bond between the carbon and nitrogen atoms, all atoms satisfy the octet rule, and the carbon attached to the nitrogen, as well as the nitrogen atom itself, have nonzero formal charges due to the triple bond and unequal sharing of electrons.
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The sun produces energy via fusion. One of the fusion reactions that occurs in the sun is 411H→42He+201e How much energy in joules is released by the fusion of 2.01 g of hydrogen-1? Express your answer to three significant figures and include the appropriate units.
Answer: The energy released for the the given amount of hydrogen -1 atom is [tex]1.2474\times 10^{11}J[/tex]
Explanation:
First we have to calculate the mass defect [tex](\Delta m)[/tex].
The given equation follows:
[tex]4_{1}^{1}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+2_0^{1}\textrm{e}[/tex]
To calculate the mass defect, we use the equation:
Mass defect = Sum of mass of product - Sum of mass of reactant
[tex]\Delta m=(2m_{e}+m_{He})-(4m_{H})[/tex]
We know that:
[tex]m_e=0.00054858g/mol\\m_{H}=1.00782g/mol\\m_{He}=4.00260g/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta m=((2\times 0.00054858)+4.00260)-(4\times 1.00782)=-0.027583g=-2.7583\times 10^{-5}kg[/tex]
(Conversion factor: 1 kg = 1000 g )
To calculate the energy released, we use Einstein equation, which is:
[tex]E=\Delta mc^2[/tex]
[tex]E=(-2.7583\times 10^{-5}kg)\times (3\times 10^8m/s)^2[/tex]
[tex]E=-2.4825\times 10^{11}J[/tex]
The energy released for 4 moles of hydrogen atom is [tex]2.4825\times 10^{11}J[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of hydrogen atom = 2.01 g
Molar mass of hydrogen atom = 1 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of hydrogen atom}=\frac{2.01g}{1g/mol}=2.01mol[/tex]
We need to calculate the energy released for the fusion of given amount of hydrogen atom. By applying unitary method, we get:
As, 4 moles of hydrogen atom releases energy of = [tex]2.4825\times 10^{11}J[/tex]
Then, 2.01 moles of hydrogen atom will release energy of = [tex]\frac{2.4825\times 10^{11}}{4}\times 2.01=1.2474\times 10^{11}J[/tex]
Hence, the energy released for the the given amount of hydrogen -1 atom is [tex]1.2474\times 10^{11}J[/tex]
How many moles of KOH are in 130.0 mL of a 0.85 M solution of KOH?
How will you known that you have reached the equivalence point when titrating the NaOH solution with HCl?
Answer:
When the solution (with phenolphthalein) changes to colorless
Explanation:
When titrating with HCl is common to add phenolphthalein as an acid-base indicator.
Phenolphthalein is pink or fucsia when added into a basic solution. On the other hand when it is in acid solutions, is colorless.
So, when titrating, the NaOH solution will be initialy pink due to the phenolphthalein and when reaching the equivalence point, that color will fade out into colorless. This is how you know you hace reached the equivalent point.
Assume that in each pair of compounds be- low the type of crystal structure is similar. Consider the strength of the attractive force
bonding the ions together in each compound.
In which case would the relative strengths be ranked correctly?
1. MgCl2 < CaBr2
2. LiCl > LiBr
3. KF < KCl
4. NaI > Li
The attraction between ions in a crystal depends on the size of the ions. Since chloride ion is smaller than bromide ion, the statement LiCl > LiBr is correct.
The attraction between oppositely charged ions in the crystal lattice of an ionic compound has a lot to do with the size of the ions.
For a given crystal lattice, smaller ions implies greater attraction and higher lattice energy.
LiCl and LiBr contains the same cation which is the lithium ion so we turn our attention to the anions.
Chloride ion is smaller than bromide ion, the statement LiCl > LiBr is correct based on the strength of the attractive force bonding the ions together in each compound .
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Final answer:
The correct ranking based on ionic bond strength due to charge and size differences is LiCl > LiBr. This considers the Coulombic attraction, where a smaller ionic radius and a higher charge increase the strength of the bond.
Explanation:
The question asks to rank the strength of the attractive force bonding the ions together in each compound pair, correctly. When considering the relative strengths of ionic bonds, factors such as ionic size and charge play significant roles. The strength of an ionic bond increases with the charge of the ions and decreases with the size of the ions. Therefore, compounds with ions of higher charges and smaller sizes have stronger bonds due to the higher Coulombic attraction.
MgCl2 < CaBr2 - Incorrect because Mg2+ has a smaller ionic radius and a higher charge density than Ca2+, leading to stronger bonding in MgCl2.
LiCl > LiBr - Correct since Cl- is smaller than Br-, leading to a stronger Coulombic attraction in LiCl.
KF < KCl - Incorrect because F- is smaller than Cl-, which should result in a stronger attraction in KF due to the smaller ionic radius of F-.
NaI > Li - This comparison is not valid as it compares an ionic compound with an element.
Therefore, the correct ranking of relative strengths based on the charge and size of ions is found in option 2: LiCl > LiBr.
Explain why the HOH molecule is bent, whereas the HBeH molecule is linear. Water, HOH, has bonding electron pairs and lone pair(s) of electrons. The placement of these electrons forces electron-pair geometry, and the resulting HOH molecule is bent. The HBeH molecule has bonding electron pairs and lone pair(s) of electrons. The bonding pairs of electrons must be as one another as possible, resulting in a linear HBeH molecule.
The question is already answered from one point of view but I'll add BeH2 has sp hybridization and H2O has sp3 hybridization getting a different structure type cause where the electron pairs are located around the central atom, that means that you will see the next structures from these molecules.
Answer:
Water (HoH) has
2
bonding electrons and 2 lone pairs. The placement of these electrons is going to be a tetrahedral electron-pair geometry. The HBeH molecule has 2 bonding electron pairs and 0 lone pairs. The bonding pairs must be as far from one another as possible.
Explanation:
I hope that helps any one in the future who needs it <3
What effect will spraying liquid water into the equilibrium given below have if NH3 is far more soluble in water than is N2 or H2?
N2(g) + 3H2(g) 2NH3(g)
A) More NH3(g) will form.
B) More H2(g) will form.
C) Less NH3(g) will form.
D) This will not affect the system.
E) More N2(g) will form.
Answer:
A) More NH3(g) will form.
Explanation:
According to the Le Chatellier Principle if we remove product from a reaction which is in equilibrium, then the effect will to produce more product in order to replace what has been removed.
In this reaction also , on spraying liquid water, Ammonia being water soluble will be consumed. Hence, The reaction will shift in forward direction leading to more formation of Ammonia (NH3).
Hence option A will be correct .
Final answer:
Spraying water into the equilibrium involving N2, H₂, and NH₃ will cause more NH₃ to form because NH₃ is more soluble in water than N₂ or H₂, and removing NH₃ by dissolving it in water shifts the equilibrium towards producing more NH₄. the correct option is A.
Explanation:
The question asks what effect spraying liquid water into the equilibrium N2(g) + 3H2(g) ⇒ 2NH3(g) will have, given that NH₃ is far more soluble in water than N2 or H2. According to Le Chatelier's Principle, if a change is applied to a reaction at equilibrium, the system will adjust to partially counteract that change. In this case, spraying water will remove NH₃ by dissolving it. This reduction in the concentration of NH3 will cause the system to shift towards the product side to replace the removed NH₃, thus more NH₃ will form. Therefore, the correct answer is A) MoreNH₃(g) will form.
Which of the following statements could be true regarding polar molecules? Choose one or more: A polar molecule will not contain polar bonds. A polar molecule may have one or more lone pairs. A polar molecule has an uneven distribution of electron density. The bond dipoles in a polar molecule will cancel. A polar molecule will never contain nonpolar bonds.
Answer:
A polar molecule may have one or more lone pairs.
A polar molecule has an uneven distribution of electron density.
Explanation:
A molecule in which the dipole bonds will not cancel is a polar molecule. It depends on the geometry of the molecule, and the direction of the dipole. So, to be polar, the molecule must have at least one polar bond. It may have nonpolar bonds, but the total dipole must be different from 0.
A polar bond is formed between elements that have large differents in electronegativities, such as chlorine and hydrogen. When an atom has a large electronegativity, it has lone pairs of electrons in a bond because it has a small size and a great number of electrons in the valence shell. So, a polar molecule may have on or more lone pairs.
Because of the lone pairs presented, and because of the dipole different from 0, it will be partial charges in the atoms, so a polar molecule has an uneven distribution of electrons density.
A polar molecule is a molecule that has opposite charges at both ends of the molecule.
The true statements about polar molecules are;
A polar molecule has an uneven distribution of electron density. A polar molecule may have one or more lone pairs.To say that a molecule is polar means that it has opposite charges at either end of the molecule.
These charges result from uneven distribution of electrons in the molecule. This uneven distribution of electrons in the molecule is also caused by the large difference in electronegativity between the bonding atoms. This electronegativity difference causes the electron density of the bond to draw closer one of bonding atoms thereby conferring a partial negative charge on that atom and a resulting partial positive charge on the other atom.
Many polar molecules have lone pairs on atoms in the molecule such as NH3, H2O, H2S, etc.
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The following reaction is found to be at equilibrium at 25 celcius: 2SO3--->O2 + 2SO2 + 198kJ/mol. If the value for Kc at 25 celcius is 8.1, are the products or reactants favored and how much?
A.) Reactants are strongly favored
B.) Reactants are weakly favored
C.) Products are strongly favored
D.) Products are weakly favored
Answer:
D.) Products are weakly favored
Explanation:
For the reaction:
2SO₃ ⇄ O₂ + 2SO₂ + 198kJ/mol
The kc is defined as:
kc = [O₂] [SO₂]² / [SO₃]²
As the kc is 8,1:
8,1 [SO₃]² = [O₂] [SO₂]²
The products are favored 8,1 times. This is a weakly favored because the usual kc are in the order of 1x10⁴. Thus, right answer is:
D.) Products are weakly favored
I hope it helps!
Final answer:
For the reaction 2SO3 → O2 + 2SO2 with an equilibrium constant Kc of 8.1 at 25 °C, the products are weakly favored since Kc is greater than 1 but much less than 10^3. So the correct answer is D.
Explanation:
The equilibrium constant Kc indicates whether the reactants or products are favored in a chemical reaction at equilibrium. For the reaction 2SO3 → O2 + 2SO2, the equilibrium constant Kc is given as 8.1 at 25 °C. Since the value of Kc is greater than 1, this means that the reaction favors the formation of products over reactants. If the Kc value was less than 1, the reactants would be favored.
According to the table provided, values of Kc greater than 103 indicate a strong tendency for the reaction to favor the formation of products, suggesting a strongly favored product side for such large Kc values. In this case, with a Kc of 8.1, which is greater than 1 but significantly less than 103, the products are favored to a lesser extent. Therefore, we could conclude that the products are weakly favored, making option D the correct answer.
A student titrated a 25.00-mL sample of a solution containing an unknown weak, diprotic acid (H2A) with N2OH. If the titration required 17.73 mL of 0.1036 M N2OH to completely neutralize the acid, calculate the concentration (in M) of the weak acid in the sample.
(a) 9.184 x 10 M
(b) 3.674 x 10-2 M
(c) 7.304 x 10-2 M
(d) 7.347 x 10-2 M
(e) 1.469 x 101 M
Answer: The concentration of weak acid is [tex]3.674\times 10^{-2}M[/tex]
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2A[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=2\\M_1=?M\\V_1=25.00mL\\n_2=1\\M_2=0.1036M\\V_2=17.73mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 25.00=1\times 0.1036\times 17.73\\\\M_1=\frac{1\times 0.1036\times 17.73}{2\times 25.00}=3.674\times 10^{-2}M[/tex]
Hence, the concentration of weak acid is [tex]3.674\times 10^{-2}M[/tex]
Consider the following reaction at constant pressure. Use the information provided below to determine the value of ΔS at 473 K. Predict whether or not this reaction will be spontaneous at this temperature.
4NH3 (g) + 3O2 (g) → 2N2 (g) + 6H2O (g) ΔH = –1267 kJ
Answer:
The reaction will be spontaneous
Explanation:
To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:
[tex]\Delta G= \Delta H - T * \Delta S [/tex]
If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.
Calculating the [tex]\Delta G= -1267 - 473 K* \Delta S [/tex] :
[tex]\Delta G= -1267 - 473 K* \Delta S [/tex]
Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.
In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in: [tex]\Delta S>0[/tex}
Back to this expression:
[tex]\Delta G= -1267 - 473 K* \Delta S [/tex]
If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.